Question

8) A proton is placed at rest near a positively charged plate. The proton is released and travels in the positive x direction. The proton reaches a velocity of 85000 m/s at the point when the potential is zero. (mproton = 1.67 x 10-27 kg, et = 1.6 x 10-19 C) a) What is the electric potential at the positively charged plate? A X b) The proton then enters a 9 T magnetic field directed into the page. Choose the picture (A or B) that accurately shows the path the proton will take. X X X B X X X X X х X х X x x x c) What is the magnitude of the force on the proton from the magnetic field? x x d) What is the radius of the path the proton will take? e) If the magnetic field were cut in half, would the radius of the path increase or decrease? Explain.

Answer 1

(8) Given, speed of the profon, d = 85000 m/ mproton = 1.67 x 10-27 ng e = 1.6x 10-19 ( @ AS , பட = utzad افا 11 0 + 2 ad اما 2x e Ed mp V = Ed → V moz - ze 1.67 x 1027 x(85000) 2 x 1.6x1079 volt = 37.7 The electric potential at the positively charged plate is 3t 7 volt The picture that accurately shows the will take, is given ty- path the proton X B. X X х X X X X X X

с on the proton The magnitude of the force on the the magnetic field is. from F e VB X 85000X9 = 1.681019 1.224 x 1013 (d) The radius of the path for the proton is given by malez evB me eB X 85000 1.67x10-27 1.6x1019 m x 9 a 9.86 x los field is cut in half If the the radius of the path magnetic inch would increase - As B gt is half then then the radius will Jouble double with deer Thus, radius incressing recensing magnetic field
Thanks